Proving the Derivative of Inverse Sine without Trigonometry

[QuarkCharmer]

Homework Statement

Prove that:
$$\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}$$

Homework Equations

The Attempt at a Solution

$$y=arcsin(x)$$
$$sin(y)=sin(arcsin(x))$$
$$sin(y)=x$$
$$\frac{d}{dx}(sin(y)=x)$$
$$cos(y)\frac{dy}{dx}=1$$
$$\frac{dy}{dx}=\frac{1}{cos(y)}$$
Using a triangle:
$$cos(y)=\sqrt{1-x^2}$$
$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$
$$\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}$$

We were shown these in class before starting trig-sub and all that, and the professor said that there are proofs by differentiation. I don't know how to explain the steps where I determine that the cosine of the angle y is equal to sqrt(1-x^2) without drawing out a triangle. Is there some way? Other than that I think I have it figured out and I can do the same for the other functions.

 

[lineintegral1]

Well, certainly,

$$x=sin(y)$$

In addition,

$$sin^2y+cos^2y=1$$

So it certainly follows that,

$$cos(y)=\sqrt{1-sin^2y}$$

You can see that it follows naturally (by substitution) that,

$$cos(y)=\sqrt{1-x^2}$$

And we're done!

 

[QuarkCharmer]

Oh haha. That really should have occurred to me considering that I used the pyth. theorem to find cos(y).

Thanks

 

[HallsofIvy]

Since you refer to "Using a triangle", you can also do it this (equivalent) way:
imagine a right triangle triangle having "opposite side" of length x and "hypotenuse" of length 1, so that sin(y)= x/1= x. Then, by the Pythgorean theorem, the "near side" has length $\sqrt{1- x^2}$. From that, $cos(y)= \sqrt{1- x^2}$.